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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jun 7, 2014 at 14:47 vote accept Ken W. Smith
Jun 7, 2014 at 14:47 vote accept Ken W. Smith
Jun 7, 2014 at 14:47
Jun 7, 2014 at 14:46 vote accept Ken W. Smith
Jun 7, 2014 at 14:47
Jun 7, 2014 at 13:42 answer added Geoff Robinson timeline score: 7
Jun 6, 2014 at 20:54 comment added Ben Webster @QiaochuYuan That normal subgroup is too small. For example, for subgroups of $SU(2)$, only cyclic and quaternion groups satisfy this condition, not the dihedrals or symmetries of the platonic solids.
Jun 6, 2014 at 20:52 answer added Ben Webster timeline score: 7
Jun 6, 2014 at 20:21 comment added Qiaochu Yuan You can use the classification of finite subgroups of $\text{Sp}(1) \cong \text{SU}(2)$, right? The normal subgroup $K$ should be the intersection of the kernels of all homomorphisms to these finite subgroups. That's not very explicit, unfortunately.
Jun 6, 2014 at 17:41 history edited Ken W. Smith CC BY-SA 3.0
added 277 characters in body
Jun 6, 2014 at 17:36 history asked Ken W. Smith CC BY-SA 3.0